I have no idea how to do it. This is very interesting problem to work on

Re: exp(5) = [$]e^5[$]

Posted: December 5th, 2018, 1:26 pm

by Cuchulainn

Wed Dec 05, 2018 1:11 pm

Let's take a step backwards.

Gradient descent methods as we know and love them are nothing more than Euler's (ugh) method applied to ODEs (aka gradient system):

[$]dx/dt = - grad f(x) = -\nabla f(x)[$] where [$]f[$] is the function to be minimised.

The local minima of [$]f[$] are the critical points of this ODE system. (Poincaré-Lyapunov-Bendixson theory).
I have tried it on a number of benchmark unconstrained optimisation problem and solved using C++ boost::odeint without any of the infamous learning rate fudges.

We use more robust ODE solvers than Euler which means that we don't get the well-documented issues with GD methods.

For the evergreen [$]e^5[$] we have ODE [$]dx/dt = -\nabla(x-e^5)^2 = -2(x - e^5)[$]. We get x = 148.413 for any initial value of the ODE.

Re: exp(5) = [$]e^5[$]

Although both [$]\pi[$] and [$]e[$] are known to be transcendental, it is not known whether the set of both of them is algebraically independent over [$]\mathbb {Q}[$]. In fact, it is not even known if [$]\pi +e[$] is irrational. Nesterenko proved in 1996 that:
the numbers [$]π[$], [$]e^{\pi}[$], and [$]\Gamma(1/4)[$] are algebraically independent over [$]\mathbb {Q}[$].
the numbers [$]π[$], [$]e^{\pi\sqrt{3}}[$], and [$]\Gamma(1/3)[$] are algebraically independent over [$]\mathbb {Q}[$].
for all positive integers [$]n[$], the numbers [$]π[$], [$]e^{\pi\sqrt{n}}[$] are algebraically independent over [$]\mathbb {Q}[$].

Re: exp(5) = [$]e^5[$]

Posted: December 15th, 2018, 12:29 pm

by ExSan

Wed Dec 05, 2018 1:11 pm

Let's take a step backwards.

Gradient descent methods as we know and love them are nothing more than Euler's (ugh) method applied to ODEs (aka gradient system):

[$]dx/dt = - grad f(x) = -\nabla f(x)[$] where [$]f[$] is the function to be minimised.

The local minima of [$]f[$] are the critical points of this ODE system. (Poincaré-Lyapunov-Bendixson theory).
I have tried it on a number of benchmark unconstrained optimisation problem and solved using C++ boost::odeint without any of the infamous learning rate fudges.

We use more robust ODE solvers than Euler which means that we don't get the well-documented issues with GD methods.

For the evergreen [$]e^5[$] we have ODE [$]dx/dt = -\nabla(x-e^5)^2 = -2(x - e^5)[$]. We get x = 148.413 for any initial value of the ODE.

Maestro !

Re: exp(5) = [$]e^5[$]

Posted: February 6th, 2019, 2:24 pm

by Cuchulainn

Does the ODE

[$]du/dt = e^{u^2/2}[$] have a soluion in which interval?

(e.g. as a first shot across boughs, maybe a symbolic solution in Maple?)

edit: added
as to the domain, you know [$]-1\le \textrm{erf}(x)\le 1[$]
so if you have an initial condition [$]u_{0}[$] at [$]t_{0}[$],
[$]\sqrt{\pi/2}\,\textrm{erf}\left(u/\sqrt{2}\right)-\sqrt{\pi/2}\,\textrm{erf}\left(u_{0}/\sqrt{2}\right)=t-t_{0}[$]
[$]t=t_{0}+\sqrt{\pi/2}\,\textrm{erf}\left(u/\sqrt{2}\right)-\sqrt{\pi/2}\,\textrm{erf}\left(u_{0}/\sqrt{2}\right)[$]
so the solution is valid for [$]t[$] in the range
[$]t_{0}-\sqrt{\pi/2}-\sqrt{\pi/2}\,\textrm{erf}\left(u_{0}/\sqrt{2}\right)\le t
\le t_{0}+\sqrt{\pi/2}-\sqrt{\pi/2}\,\textrm{erf}\left(u_{0}/\sqrt{2}\right)[$]
or in terms of [$]\textrm{erfc}(x)=1-\textrm{erf}(x)[$] which is between 0 and 2
[$]t_{0}-\sqrt{\pi/2}\,\textrm{erfc}\left(-u_{0}/\sqrt{2}\right)\le t
\le t_{0}+\sqrt{\pi/2}\,\textrm{erfc}\left(u_{0}/\sqrt{2}\right)[$]

a very rough limit would be to replace erfc by 2,
[$]t_{0}-\sqrt{2\pi}\le t\le t_{0}+\sqrt{2\pi}[$]

Re: exp(5) = [$]e^5[$]

Posted: February 9th, 2019, 10:01 pm

by ExSan

Does the ODE

[$]du/dt = e^{u^2/2}[$] have a soluion in which interval?

(e.g. as a first shot across boughs, maybe a symbolic solution in Maple?)

edit: added
as to the domain, you know [$]-1\le \textrm{erf}(x)\le 1[$]
so if you have an initial condition [$]u_{0}[$] at [$]t_{0}[$],
[$]\sqrt{\pi/2}\,\textrm{erf}\left(u/\sqrt{2}\right)-\sqrt{\pi/2}\,\textrm{erf}\left(u_{0}/\sqrt{2}\right)=t-t_{0}[$]
[$]t=t_{0}+\sqrt{\pi/2}\,\textrm{erf}\left(u/\sqrt{2}\right)-\sqrt{\pi/2}\,\textrm{erf}\left(u_{0}/\sqrt{2}\right)[$]
so the solution is valid for [$]t[$] in the range
[$]t_{0}-\sqrt{\pi/2}-\sqrt{\pi/2}\,\textrm{erf}\left(u_{0}/\sqrt{2}\right)\le t
\le t_{0}+\sqrt{\pi/2}-\sqrt{\pi/2}\,\textrm{erf}\left(u_{0}/\sqrt{2}\right)[$]
or in terms of [$]\textrm{erfc}(x)=1-\textrm{erf}(x)[$] which is between 0 and 2
[$]t_{0}-\sqrt{\pi/2}\,\textrm{erfc}\left(-u_{0}/\sqrt{2}\right)\le t
\le t_{0}+\sqrt{\pi/2}\,\textrm{erfc}\left(u_{0}/\sqrt{2}\right)[$]

a very rough limit would be to replace erfc by 2,
[$]t_{0}-\sqrt{2\pi}\le t\le t_{0}+\sqrt{2\pi}[$]